Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include"bits/stdc++.h"

using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

#define FOR(k,m,n) for(ll (k)=(m);(k)<(n);(k)++)
#define REP(i,n) FOR((i),0,(n))
#define WAITING(str) int str;std::cin>>str;
#define DEBUGING(str) cout<< #str << " " str<<endl

constexpr int INF = (1 << 30);
constexpr ll INFL = (1ll << 60);
constexpr ll MOD = 1000000007;// 10^9+7

int N, M, s, t;
vector<vector<int>> edges;

vector<int> dijkstra(int s) {
	priority_queue<pii, vector<pii>, greater<pii>> pq;
	vector<int> d(N, INF);

	d[s] = 0;
	pq.push({ 0,s });
	while (!pq.empty()) {
		auto now = pq.top(); pq.pop();
		int v = now.second;
		for (auto next : edges[v]) {
			if (d[next] > d[v] + 1) {
				d[next] = d[v] + 1;
				pq.push({ d[next],next });
			}
		}
	}
	return d;
}

int main()
{
	cin >> N >> M >> s >> t;
	s--; t--;
	edges.resize(N);
	REP(i, M) {
		int x, y;
		cin >> x >> y;
		x--; y--;
		edges[x].push_back(y);
		edges[y].push_back(x);
	}
	auto dl = dijkstra(s);
	auto dr = dijkstra(t);
	int len = dl[t];

	// len - cnt
	map<int, int> mpl, mpr;
	for (auto d : dl)mpl[d]++;
	for (auto d : dr)mpr[d]++;

	ll res = 0;
	REP(i, len - 1) {
		int l = i;
		int r = len - 2 - i;
		if (mpl.find(l) == mpl.end())continue;
		if (mpr.find(r) == mpr.end())continue;
		res += (ll)mpl[l] * mpr[r];
	}
	cout << res << endl;

	return 0;
}